# Question 90093

Mar 16, 2017

$- \text{719 kJ}$

#### Explanation:

Before moving on, it's worth mentioning that the actual value for the standard enthalpy of combustion of butane is

$\Delta {H}_{\text{comb"^@ = - "2877.5 kJ mol}}^{- 1}$

I'll use the actual value in my calculations, but you can replace it with the value given to you if needed.

So, the first thing to do here is to convert the standard enthalpy of combustion of butane from kilojoules per mole to kilojoules per gram.

The molar mass of butane can be calculated by using

M_ ("M C"_ 4"H"_ 10) = 4 xx M_ "M C" + 10 xx M_ "M H"

This will get you

M_ ("M C" _ 4"H"_ 10) = 4 xx "12 g mol"^(-1) + 10 xx "1 g mol"^(-1)

M_ ("M C" _ 4"H"_ 10) = "58 g mol"^(-1)#

Now, you know that the standard enthalpy of combustion of butane is equal to

$\Delta {H}_{\text{comb"^@ = - "2877.5 kJ mol}}^{- 1}$

This tells you that when $1$ mole of butane undergoes combustion, $\text{2877.5 kJ}$ of heat are being given off, hence the minus sign.

Use the molar mass of the compound to convert this to kilojoules per gram

$- {\text{2877.5 kJ"/(1 color(red)(cancel(color(black)("mole")))) * (1color(red)(cancel(color(black)("mole C"_4"H"_10))))/("58 g") = -"49.612 kJ g}}^{- 1}$

This tells you that when $\text{1 g}$ of butane undergoes combustion, $\text{49.612 kJ}$ of heat are being given off.

You can thus say that the enthalpy change of combustion for your sample will be equal to

$14.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * (-"49.612 kJ")/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(-"719 kJ}}}}$

The answer is rounded to three sig figs.